Seven
First rewrite the original number as 10t+u where u represents the broken off digit and t is the other new number. So the original number is 10t+u. The number created by subtracting "(p-n)×u", where (p-n)=2, from the other number gives t-2u. The claim of the test is that if 7|(t-2u) then 7|(10t+u). So by manipulating the equation by multiplying by numbers relatively prime to 10 and adding multiples of 7 (both of which don't change divisibility):
7|(t-2u) if and only if
7|10(t-2u) or 7|(10t-20u) if and only if
7|(10t-20u+21u) or 7|(10t+u).
Therefore 7|(t-2u) if and only if 7|(10t+u) and vice versa.
Eleven
Again rewrite the original number as 10t+u where u represents the broken off digit and t is the other new number. So the original number is 10t+u. The number created by subtracting "(p-n)×u", where (p-n)=1, from the other number gives t-u. The claim of the test is that if 11|(t-u) then 11|(10t+u). So, again,by manipulating the equation by multiplying by numbers relatively prime to 10 and adding multiples of 11:
11|(t-u) if and only if
11|10(t-u) or 11|(10t-10u) if and only if
11|(10t-10u+11u) or 11|(10t+u).
Therefore 11|(t-u) if and only if 11|(10t+u) and vice versa.
Thirteen
Again rewrite the original number as 10t+u where u represents the broken off digit and t is the other new number. So the original number is 10t+u. The number created by adding "n×u", where n=4, from the other number gives t+4u. The claim of the test this time is that if 13|(t+4u) then 13|(10t+u). So by manipulating the equation by multiplying by numbers relatively prime to 10 and subtracting multiples of 13:
13|(t+4u) if and only if
13|10(t+4u) or 13|(10t+40u) if and only if
13|(10t+40u-39u) or 13|(10t+u).
Therefore 13|(t+4u) if and only if 13|(10t+u) and vice versa.
...As you can see the proof becomes repetitive and it can easily be adapted to other primes.